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ollegr [7]
2 years ago
10

This is for AP CALCULUS. I really need help on this. I’m trying to find the first and second derivative of these three functions

.

Mathematics
1 answer:
ale4655 [162]2 years ago
8 0

Review Material:

\sqrt[n]{x^m} = x^{\frac{m}{n}}

\frac{d}{dx}[x^n]=nx^{n-1}\\

\frac{d}{dx}[\sin(x)]=\cos(x)\\

 \frac{d}{dx}[\cos(x)]=-\sin(x)\\

\frac{d}{dx}[constant]=0\\

Step-by-step explanation:

(a)

F(x) = -2\cos(x) +x^{\frac{4}{3}} -3e\\ \text{Note: 3e is a constant}\\F'(x) = 2\sin(x) + \frac{4}{3}x^{\frac{1}{3}}\\F''(x) = 2\cos(x) + \frac{4}{9}x^{-\frac{2}{3}}

(b)

F(x) = x^{-3} + \frac{1}{2}x^2-\sin(x)\\F'(x)=-3x^{-4} + x - \cos(x)\\F''(x) = 12x^{-5} + 1 + \sin(x)

(c)

F(x) = 2x^{-3}+x^{\frac{3}{4}}-4x\\F'(x) = -6x^{-4} + \frac{3}{4}x^{-\frac{1}{4}}-4\\F''(x)= 24x^{-5} - \frac{3}{16}x^{-\frac{5}{4}}

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Find a recurrence relation for the amount of money in a savings account after n years if the interest rate is 6 percent and $50
Korvikt [17]

Answer:

a_n = a_{n-1} (1.06) + 50

Step-by-step explanation:

Suppose, a_0 is initial amount in the saving account,

Here, the annual interest rate is 6% and additional amount in each year is $ 50,

So, the amount after one year,

a_1 = a_0 + 6\%\text{ of }a_0 + 50 = a_0 + 0.06a_0 + 50 = a_0(1.06) + 50

Amount after 2 years,

a_2 = a_1 + 6\%\text{ of }a_1 + 50 = a_1(1.06) + 50

Amount after 3 years,

a_3 = a_2 + 6\%\text{ of }a_2 + 50 = a_2(1.06) + 50

................................., so on....

Hence, by following the pattern,

The amount after n years,

a_n = a_{n-1} (1.06) + 50

Which is the required recurrence relation for the amount of money in a savings account

3 0
3 years ago
Please help, I don't know how to do this and need it done quickly, thanks for your help and time! have a good day :)
Olenka [21]
Sin50° = BC/3
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7 0
3 years ago
Which equation is equivalent to log Subscript 3 Baseline (x + 5) = 2?
adell [148]

Answer:

The correct option is right-bracket squared 3 squared =x+5

Step-by-step explanation:

The equation is \log _{3}(x+5)=2

Option a: \log _{3}(x+5)=3^{2}

This is not possible because using logarithmic rule, if \log _{a} b=c then b=a^{c}

Hence, option a is not equivalent to \log _{3}(x+5)=2

Option b: \log _{3}(x+5)=2^{3}

This is not possible because using logarithmic rule, if \log _{a} b=c then b=a^{c}

Hence, option b is not equivalent to \log _{3}(x+5)=2

Option c: x+5=3^{2}

This is possible because using logarithmic rule, if \log _{a} b=c then b=a^{c}

Hence, option c is equivalent to \log _{3}(x+5)=2

Option b: x+5=2^{3}

This is not possible because using logarithmic rule, if \log _{a} b=c then b=a^{c}

Hence, option b is not equivalent to \log _{3}(x+5)=2

Thus, the correct option is c: x+5=3^{2}

Hence, the equation x+5=3^{2} is equivalent to \log _{3}(x+5)=2

8 0
3 years ago
A number is greater than 8. The same number is less than 10. The inequalities x greater-than 8 and x less-than 10 represent the
jok3333 [9.3K]

Answer:

There are a few solutions because there are some fractions and decimals between 8 and 10

Step-by-step explanation:

Let the unknown number be 'x'

If the number is greater than 8 and the same number is less than 10, this can be expressed as;

x>8 and x < 10

Note that if x>8, then 8<x

The resulting inequalities are now;

8<x and x<10

Combining both inequalities we have: 8<x<10

Since the inequality didn't tell us that the variable 'x' is equal to 8 and 10, this means that our solution falls between 8 and 10 and the value of integer that falls within this range is 9. Other values that falls within this range are decimals and fractions.

Therefore it can be concluded that there are a few solutions because there are some fractions and decimals between 8 and 10

4 0
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Answer:

Radiant energy is the energy of electromagnetic waves.

Step-by-step explanation:

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3 years ago
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